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CTK Exchange
Henry Heck (Guest)
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Dec-17-00, 11:23 AM (EST) |
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"Poisson processes"
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Thank you for your wonderful mathematical website! I have encountered a problem in my research to which (it'seems to me) there are two possible solutions, and I don't know which (if either) is correct. Can you suggest an opinion? Without going into great detail, the problem is this. Suppose that you have a simple Poisson process that is dependent on time, i.e., state 0 -> state 1 -> state 2 -> etc., so that the parameter lt in the Poisson formula Pn(t) = (lt){n}e{-(lt)}/n! (n = 0, 1, 2, 3, etc.; exponential values signified by brackets, e.g., both {n} and {-(lt)} are exponents) increases linearly with time. The symbol -> means "evolves to" or "changes to", where the rate constant of evolution or change is l, and the same rate constant characterizes each step in the process. Let us denote this simple Poisson process as the "primary process". Now suppose that there is a secondary process that is also linearly dependent on time, but that the rate of this secondary process depends on the state of the primary (Poisson) process. In the particular case of interest, the rate of the secondary process decreases as the state of the Poisson process increases from state 0 to state 1 to state 2, etc. In other words, as the Poisson process "matures" over time, the secondary process appears to proceed more and more slowly. The particular example that concerns me is DNA base modifications (primary process) and their effects on DNA replication (secondary process). As DNA base modifications increase over time (due to exposure to UV light or to mutagens), the apparent rate of DNA replication decreases, because DNA base modifications block DNA replication. Therefore, the length of DNA that is replicated in the presence of base modifications in a given time T is shorter than the length of DNA that would be replicated in the absence of base modifications. Thus, the primary process inhibits the secondary process. The question is: By what fraction has the extent of the secondary process been altered by the primary process, when both processes occur simultaneously and linearly with time? One possible solution is as follows. (1) Calculate the time-averaged value of the primary process over the interval 0 -> T. This value is obviously lT/2. (2) Substitute lT/2 into the Poisson formula and calculate the time-averaged distribution of states. (3) From the time-averaged distribution of states (each of which has a characteristic rate for the secondary process), calculate the "average" extent of the secondary process at time T. The second possible solution is as follows. (1) Treat lt as a variable. (2) Substitute lt into the Poisson formula to calculate the distribution of states and the rate of the secondary process at time t. (3) Using the Poisson formula to determine the distribution of states, integrate over time (0 -> T) to obtain the extent of the secondary process at time T. My personal opinion is that the first solution is the more correct one. The first solution considers the distribution of states at time T/2 (average time - a fixed quantity), whereas definite integration means that the second solution emphasizes the distribution of states at time T (final time - also a fixed quantity). The first solution is a simple algebraic one, whereas the second solution involves some elementary calculus. Do you have an opinion on the proper route to follow? Thanking you in advance for your advice, - Henry Heck, Ph.D. |
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alexb
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Dec-17-00, 11:31 AM (EST) |
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1. "RE: Poisson processes"
In response to message #0
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Henry, hello: Thank you for the very kind words. I do not know of how much help I may be. My knowledge of stochastic processes is quite superficial. One thing though I am certain of is that the average value of a Poisson process over an interval of length T is lT, and not lT/2. This you obtain by a regular summation sum(nPn(t)), which is simplified by reindexing n=n+1, with the net result of the factor lt. So, it may be that both approaches in fact lead to about the same estimation, perhaps with different degrees of accuracy. All the best, Alexander Bogomolny |
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